Lemma 7.5(Katznelson's book on harmonic analysis) A series $S\sim\sum a_ne^{int}$ is the Fourier-Stieltjes series of a positive measure if, and only if, for all $n$ and $t\in\mathbb T$, $$ \sigma_n(S,t)=\sum_{-n}^n(1-|j|/(n+1))a_je^{ijt}\ge 0. $$ The proof is rather strightforward, but the author remarked that the condition "$\sigma_n(S,t)\ge 0$ for all $n$" can clearly be replaced by "$\sigma_n(S,t)\ge 0$ for infinitely many $n$'s".
I cannot figure out the reason why we can weaken the premise.
Note that we have $\sigma_n(S,t)\ge 0$ for infinitely many $n$ such that $$ \|\sigma_n(S)\|_{M(\mathbb T)}=\frac 1{2\pi}\int\sigma_n(S,t)dt=a_0 .$$ So in order to use Corollary 7.3, we shall try to argue $\|\sigma_n(S)\|\le C$ for all $n$ where $C$ is a constant. Any suggestion?
The same lemma is also discussed under:Fourier Stieltjes series of a positive measure but I have a different question.
Addendum:
Proof of Lemma 7.5:
If $S=S(\mu)$ for a positive $\mu\in M(\mathbb T)$ and if $f\in C(\mathbb T)$ is non-negative, we have $$ \frac{1}{2\pi}\int f(t)\overline{\sigma_n (S,t)}dt=\sum_{-n}^n\left( 1-\frac{|j|}{n+1} \right)\hat f(j)\overline{\hat\mu(j)}=\int\sigma_n(f)\overline{d\mu}\ge 0 $$ since $\mu\ge 0$ and, by 3.1, $\sigma_n(f,t)>0$. Since this is true for arbitrary nonnegative $f$, $\sigma_n(S,t)\ge 0$ on $\mathbb T$. Assuming $\sigma_n(S,t)\ge 0$ we obtain $$ \|\sigma_n(S)\|_{M(\mathbb T)}=\frac 1{2\pi}\int \sigma_n(S,t)dt=a_0 $$ and, by Corollary 7.3, $S=S(\mu)$ for some $\mu\in M(\mathbb T)$. For arbitrary nonnegative $f\in C(\mathbb T)$, $\int f d\mu=\lim_{n\to\infty}1/2\pi \int f(t)\sigma_n(S,t) dt\ge 0$ and it follows that $\mu$ is a positive measure. Q.E.D.